Question: Solve for $x$ : $ 4|x + 2| - 7 = 1|x + 2| + 5 $
Subtract $ {1|x + 2|} $ from both sides: $ \begin{eqnarray} 4|x + 2| - 7 &=& 1|x + 2| + 5 \\ \\ { - 1|x + 2|} && { - 1|x + 2|} \\ \\ 3|x + 2| - 7 &=& 5 \end{eqnarray} $ Add ${7}$ to both sides: $ \begin{eqnarray} 3|x + 2| - 7 &=& 5 \\ \\ { + 7} &=& { + 7} \\ \\ 3|x + 2| &=& 12 \end{eqnarray} $ Divide both sides by ${3}$ $ \dfrac{3|x + 2|} {{3}} = \dfrac{12} {{3}} $ Simplify: $ |x + 2| = 4$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x + 2 = -4 $ or $ x + 2 = 4 $ Solve for the solution where $x + 2$ is negative: $ x + 2 = -4 $ Subtract ${2}$ from both sides: $ \begin{eqnarray} x + 2 &=& -4 \\ \\ {- 2} && {- 2} \\ \\ x &=& -4 - 2 \end{eqnarray} $ $ x = -6 $ Then calculate the solution where $x + 2$ is positive: $ x + 2 = 4 $ Subtract ${2}$ from both sides: $ \begin{eqnarray} x + 2 &=& 4 \\ \\ {- 2} && {- 2} \\ \\ x &=& 4 - 2 \end{eqnarray} $ $ x = 2 $ Thus, the correct answer is $x = -6 $ or $x = 2 $.